A wave on a rope is described by y = 0.27cos(2.69 x + 1.25* t ), where y and x

ID: 2288592 • Letter: A

Question

A wave on a rope is described by y = 0.27*cos(2.69*x + 1.25*t), where y and x are in meters and t is in seconds.

a.) Determine the wavelength. (include units with answer)

b.) Determine the frequency. (include units with answer)

c.) Determine the maximum amplitude. (include units with answer)

d.) Determine the wave velocity. (include units with answer)

e.) Determine the maximum velocity of the rope. (include units with answer)

f.) Determine the maximum acceleration of the rope.(g)

I will rate. please just show how to work so i can understand.

A wave on a rope is described by y = 0.27*cos(2.69*x + 1.25*t)

This is in the form of y =Acos(kx+wt) comparing both the equations we get

We know that k =2pi/lamda

Then lamda= 2pi/k =2*3.14/2.69 =2.33m

We know that w =2pif then f =w/2pi =1.25/2*3.14 =0.199Hz

The maxiumum amplitude is given by A =0.27m above equation

The wave velcoty is given by v =f*lamda =(0.199)(2.33) =0.463m/s

The maximum veocity is vmax =Aw =A*2pif =(0.27)(2*3.14*0.199) =0.337m/s

The maximum acceleration is given by amax =Aw2= (0.27)(2*3.14*0.199)2 =0.421m/s2